Question

Find the net electric potential at the point $\text{A}$ for the figure shown below :

• A. $\frac{Kq}{5}$
• B. $\frac{Kq}{4}$
• C. $Kq$
• D. $\frac{Kq}{2}$

Answer 1

The correct option is A $\frac{Kq}{5}$

Charge induced on the conductor due to induction is shown below.



As we know, conductors are equipotentials. Since a charge is free to move around in a conductor, no work is done in moving a charge from one point in a conductor to another. That means the electric potential inside the conductor is constant.

So, potential at point $\text{A}$ will be equal to the potential at point $\text{O}$.
i.e. $V_A=V_O$

Potential at point $\text{O}$ is the sum of potentials due to charge $q$ and due to induced charge on the conductor.

$V_O=V_q+V_{ind}$

Due to symmetry, $V_{ind}=0$
[charges are equidistant from point $\text{O}$ and there are equal no. of positive and negative charges]

$\Rightarrow V_O=\frac{Kq}{r}+0=\frac{Kq}{5}$

So, $V_A=V_O=\frac{Kq}{5}$

Hence, option (b) is the correct answer.